为什么下列代码(简单的C语言计算器代码)在通过编译和组建后却无法正确运行? C语言编简单计算器,我程序哪里出错了,为什么运行出来2+5是...

\u4e3a\u4ec0\u4e48C\u8bed\u8a00\u7a0b\u5e8f\u7f16\u8bd1\u6b63\u786e\u4f46\u4e0d\u80fd\u6267\u884c\uff1f

\u8fd9\u5f53\u7136\u539f\u56e0\u975e\u5e38\u591a\uff0c\u6bd4\u5982\uff1a\u7f16\u8bd1\u5668\u7f16\u8bd1\u4e0d\u591f\u4e25\u683c\u3001\u7a0b\u5e8f\u6709\u5f88\u5927\u7684\u98ce\u9669\u7b49\u7b49\uff0c\u6700\u7b80\u5355\u7684\u4e00\u4e2a\u4f8b\u5b50\u5982\u4e0b\uff1a
int a; scanf("%d",a);\u53ef\u4ee5\u786e\u5b9a\u7684\u4e00\u70b9\u662f\uff1a\u4e00\u65e6\u5728\u8fd0\u884c\u4e2d\u51fa\u73b0\u9519\u8bef\uff0c\u5927\u591a\u662f\u8f83\u4e3a\u4e25\u91cd\u7684\u9519\u8bef\uff0c\u987b\u5f15\u8d77\u91cd\u89c6\uff0c\u5426\u5219\uff0c\u4e00\u65e6\u9519\u8bef\u975e\u5e38\u4e25\u91cd\uff0c\u5fc5\u4f1a\u5bfc\u81f4\u8bb8\u591a\u4e25\u91cd\u7684\u540e\u679c\uff0c\u5982\u6b7b\u673a\u3001\u5361\u5c4f\u751a\u81f3\u76f4\u63a5\u5bfc\u81f4\u84dd\u5c4f\u3002

\u4f60\u628a\u5224\u65ad\u4e2d\u7684\u53cc\u5f15\u53f7\u6362\u6210\u5355\u5f15\u53f7\uff0c\u5e94\u8be5\u5c31\u5bf9\u4e86\uff01op\u662f\u5b57\u7b26\u578b\u53d8\u91cf\uff0c\u4e0d\u662f\u5b57\u7b26\u578b\u6570\u7ec4\uff0c\u6240\u4ee5\u5e94\u8be5\u7528\u5355\u5f15\u53f7\u800c\u4e0d\u662f\u53cc\u5f15\u53f7

很明显,你的代码在处理1+2+3的case上存在bug的,而处理(1+2)+3或者1+(2+3)则是OK的,查看错误的原因也是if(o=='\n')不成立而走入了else无法解析字符+

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